Question

In a triangle ABC, if cos A = 4/5, cos B = 3/5 find a:b:c

Answer 1

For the given triangle ABC, the ratio of its sides, a:b:c will be 3:4:5.

Given,

In a ΔABC,

$\cos A=\frac{4}{5} ,$

$\cos B=\frac{3}{5}.$

To find,

The ratio of its sides, a:b:c.

Solution,

It can be seen that here, the cosine of two angles A and B, of a triangle ABC are given, as follows.

$\cos A=\frac{4}{5} ,$

$\cos B=\frac{3}{5}.$

Now, as we know that the cosine of an angle in a right triangle is given as,

$\cos \theta=\frac{base}{hypotenuse}$

Thus, when two different angles A and B, as given are considered, we can see,

$\cos A =\frac{base}{hypotenuse}=\frac{4}{5} ,$

$\cos B =\frac{base}{hypotenuse}=\frac{3}{5}.$

Thus, the triangle formed will be as shown in the figure.

As we can see from the figure, the sides of the triangle ABC will be in the ratio,

a:b:c = 3:4:5.

Therefore, for the given triangle ABC, the ratio of its sides, a:b:c will be 3:4:5.

#SPJ3

Answer 2

Answer:

a:b:c = 100/36

Step-by-step explanation:

In Triangle ABC

cosA=4/5

i.e. AC/AB=4/5

→AB=5/4×AC------(1)

Now

cosB=3/5

i.e. BC/AB=3/5

→AB=5/3×BC-------(2)

from equation (1)&(2)

5/4×AC=5/3×BC

→AC/BC=4/3

A:B:C=AC/AB×BC/AB×AC/BC

=5/4×5/3×4/3

=100/36 Ans

#SPJ3